Question

At least one of the answers above is NOT correct. (1 point) Suppose f(x, t) = e 3t sin(x + 2t). (a) At any point (x, t), the differential is df = e^(-3t)cos(x+2t)dx+(e^(-3t))(2cos(x+2t)-2sin(x+2t))dt (b) At the point (-1,0), the differential is df = cos(-1)dx+(2cos(-1))+3sin(-1)dt (c) At the point (-1,0) with dx = -0.5 and dt = 0.3, the differential is df = 0.97344 Note. You can earn partial credit on this nrohlem

Answer 1

a) Here given franssa e 3t fin (4426) First note that .. pa (isin (0424) = COJ (2+273 (34249 - cos 1872t).' = coscat ry and al Sindf2t)) = cos (2424) & (427) = 2 LOJCAH? and at (e-34) =-3 e 3t Now an ta feasts, hence - df = of in de tot adt ory of = 0 [e-31 in (at2t] da to [est, in (04-24)] dr = é 3t cos (atzt) da +C -34-34fon (+2434zē cetudo 1

--.f- testos(2+24) date }(2.cos(a+z+) =3 fin(x+4)dt At the point (-1,0) (rea.) Page 1 l} = eº ca (-) 4 + 4( 2 cm (-) - 36 - 4 1 or, if = eos(-1) dx + (2 cos (f 1)-3 sin coldt © At point (-1,0) with (rea) - da = -0.5 and dt = 0.3

the differential in df = 7-6015) Cos (41) +0.3(2005615–3 Sinev) Crean Tpulting da=-0.5, dt= 0.3